Discrete complex reflection groups

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 12 May 2014

Notes and References

This is an excerpt of the lecture notes Discrete complex reflection groups by V.L. Popov. Lectures delivered at the Mathematical Institute, Rijksuniversiteit Utrecht, October 1980.

The structure of $r\text{-groups}$ in the case $s=n+1$

When $s=n+1$ it is no longer true in general that an infinite irreducible complex crystallograph $r\text{-group}$ $W$ is the semidirect product of $\text{Lin}\hspace{0.17em}W$ and $\text{Tran}\hspace{0.17em}W\text{.}$ We shall explain here how one can find the corresponding extensions of $\text{Tran}\hspace{0.17em}W$ by $\text{Lin}\hspace{0.17em}W$ in this case.

We assume that $k=ℂ\text{.}$

The Cocycle $c$

We recall that the structure of the extension is given by a $1\text{-cocycle}$ $c$ of $\stackrel{\sim }{K},$ where $K=\text{Lin}\hspace{0.17em}W,$ with values in $V\text{;}$ see Section 2.4.

Let $\mathrm{\Gamma }=\text{Tran}\hspace{0.17em}W$ and $a\in E\text{.}$ Taking $a$ as an origin, we can identify $A\left(E\right)$ and $GL\left(V\right)V\text{;}$ then $W$ consists of the elements $(P,c\left(P\right)+\mathrm{\Gamma }),$ $P\in K\text{.}$

First of all, one can assume, upon replacing $c$ by a suitable cocycle homologuous to $c,$ that $$c\left(r_1\right)=\dots =c\left(r_n\right)=0\text{.}$$ We assume that the order of $R_j$ is equal to $m\left(H_{R_j}\right),$ $1\le j\le s,$ see Section 3.2.

		Proof.
	Let ${\gamma }_1,\dots ,{\gamma }_n\in W$ be the reflections with $\text{Lin}\hspace{0.17em}{\gamma }_j=R_j,$ $1\le j\le n\text{.}$ We assume, as usual, that the group $K\prime$ generated by $R_j,$ $1\le j\le n,$ is irreducible. We have $\underset{j=1}{\overset{n}{\cap }}{H}_{R_j}=0,$ hence $\underset{j=1}{\overset{n}{\cap }}{H}_{{\gamma }_j}=b\in E\text{.}$ Then we have $${\kappa }_b\left({\gamma }_j\right)=(R_j,0),\hspace{0.17em}j=1,\dots ,n,$$ and we are done. $\square$

We shall assume now that $$c\left(r_1\right)=\dots =c\left(r_n\right)=0\text{.}$$

Therefore $c$ is defined by only one vector $c\left(r_{n+1}\right)\text{.}$ Moreover, one can take $$c\left(r_{n+1}\right)=\lambda e_{n+1},\hspace{0.17em}\lambda \in ℂ$$ because there exists a reflection $(R_{n+1},v)$ in $W\text{.}$

Therefore, given an invariant lattice $\mathrm{\Gamma }$ of full rank, one has to solve the following problems

Find those $\lambda \in ℂ$ for which

a)	the cocycle $c$ of $\stackrel{\sim }{K}$ with values in $V,$ given by condition $$c\left(r_j\right)=0,\hspace{0.17em}1\le j\le n\text{;}\hspace{0.17em}c\left(r_{n+1}\right)=\lambda e_{n+1},\hspace{0.17em}\lambda \in ℂ,$$ satisfies $$c\left(F\right)\in \mathrm{\Gamma }$$ for any relation $F$ of $K$ (i.e. for every element $F$ of $\text{Ker}\hspace{0.17em}\varphi ,$ see Section 2.4).
b)	Condition a) holds and the corresponding group $W$ is an $r\text{-group.}$

Theorem.

1)	If $\mathrm{\Gamma }={\mathrm{\Gamma }}^0$ then a) $\Rightarrow$ b).
2)	If $c\left(F\right)\in {\mathrm{\Gamma }}^0$ for every relation $F$ then b) $\Rightarrow$ $\mathrm{\Gamma }={\mathrm{\Gamma }}^0\text{.}$

		Proof.
	1) Let $\mathrm{\Gamma }={\mathrm{\Gamma }}^0\text{.}$ We know that $${\mathrm{\Gamma }}^0={\mathrm{\Gamma }}_1+\dots +{\mathrm{\Gamma }}_{n+1}\text{.}$$ But $\mathrm{\Gamma }\prime ={\mathrm{\Gamma }}_1+\dots +{\mathrm{\Gamma }}_n$ is a root lattice for $K\prime$ and the condition $c\left(r_1\right)=\dots =c\left(r_n\right)=0$ shows that the semidirect product of $K\prime$ and $\mathrm{\Gamma }\prime$ lies in $W\text{.}$ This semidirect product is an $r\text{-group,}$ see Section 4.5. We also have that $W$ contains the reflections $(R_{n+1},\lambda e_{n+1}+t),$ $t\in {\mathrm{\Gamma }}_{n+1}\text{.}$ Let $\gamma \in W$ be an arbitrary element, $\gamma =(P,v)\text{.}$ Then $P$ is a product of reflections $R_j,$ $1\le j\le n+1,$ in some order. Therefore, multiplying $\gamma$ by $(R_j,0),$ $1\le j\le n,$ and by $(R_{n+1},\lambda e_{n+1})$ in a suitable order, one can obtain an element of the form $(1,t)\text{.}$ But the elements $(1,t\prime ),$ $t\prime \in \mathrm{\Gamma }\prime ,$ are also products of reflections. Therefore, multiplying $\gamma$ by reflections, one can obtain $(1,r),$ $r\in {\mathrm{\Gamma }}_{n+1}\text{.}$ But $$(R_{n+1},\lambda e_{n+1})(R_{n+1},\lambda e_{n+1}+r)(1,r)=(1,0)\text{.}$$ Therefore $\gamma$ is a product of reflections. 2) We have $c\left(F\right)\in {\mathrm{\Gamma }}^0\text{.}$ Therefore $c,$ in fact, defines a $1\text{-cocycle}$ of $K$ with values in $V/{\mathrm{\Gamma }}^0\text{.}$ Let $W\prime$ be the group defined by $c,$ with $\text{Lin}\hspace{0.17em}W\prime =K$ and $\text{Tran}\hspace{0.17em}W\prime ={\mathrm{\Gamma }}^0\text{.}$ It is an $r\text{-group}$ because of 1). Also we have a group $W$ defined by $c,$ with $\text{Lin}\hspace{0.17em}W=K$ and $\text{Tran}\hspace{0.17em}W=\mathrm{\Gamma }\text{.}$ Let $\gamma \in W$ be a reflection. By section 3.2, we know, there exists $\delta \in W\prime$ with $\delta \gamma {\delta }^{-1}=(R_j^l,t)$ for certain $l$ and $j\text{.}$ This is also a reflection, hence $t\perp H_{R_j}\text{.}$ But $t=c\left(r_j^l\right)+v,$ $v\in \mathrm{\Gamma }\text{.}$ We have $$c\left(r_j^l\right)=(1+r_j+r_j^2+\dots +r_j^{l-1})c\left(r_j\right)\text{.}$$ By definition of $c,$ we have $c\left(r_j\right)\perp H_{R_j}\text{.}$ Hence, also $c\left(r_j^l\right)\perp H_{R_j}\text{.}$ Therefore, $v\perp H_{R_j},$ or, in other words, $v\in {\mathrm{\Gamma }}^0\text{.}$ This means that $\delta \gamma {\delta }^{-1}\in W\prime \text{.}$ It follows now from $\delta \in W\prime$ that $\gamma \in W\prime \text{;}$ hence $W=W\prime$ and $\mathrm{\Gamma }={\mathrm{\Gamma }}^0\text{.}$ $\square$

The following simple observation is very useful in practice because it gives strong restrictions on the choice of $\lambda \text{:}$

Theorem. Let a) be satisfied and let $P\in K\prime$ be such that $$R_{n+1}P{R}_{n+1}^{-1}\in K\prime \text{.}$$ Then $$\lambda (1-R_{n+1}P{R}_{n+1}^{-1})e_{n+1}\in \mathrm{\Gamma }\text{.}$$

		Proof.
	It follows from the definition of $c$ that $(P,0)$ and $(R_{n+1},\lambda e_{n+1})\in W\text{.}$ We have now: $$(R_{n+1},\lambda e_{n+1})(P,0){(R_{n+1},\lambda e_{n+1})}^{-1}=(R_{n+1}P{R}_{n+1}^{-1},-R_{n+1}P{R}_{n+1}^{-1}\left(\lambda e_{n+1}\right)+\lambda e_{n+1})$$ and we are done. $\square$

Example

$K=K_{31}\text{.}$ The graph is $$\begin{array}{c} 1 2 2 2 2 3 4 5 i i-1 \end{array}$$ The vectors $e_1,e_2,\dots ,e_5$ are given in Table 2. Note that $e_5=i{e}_1+e_2+e_3\text{.}$

There exists only one (up to similarity) $K\text{-invariant}$ lattice $\mathrm{\Gamma }$ of full rank $$\mathrm{\Gamma }=[1,i]e_1+\dots +[1,i]e_4\text{.}$$

The presentation (i.e. the sets of generators and relations) of $K$ is known, see [Cox1974].

The relations are: $$\begin{array}{c}{R}_1^2=R_2^2=R_3^2={\left(R_2{R}_3\right)}^3={\left(R_3{R}_1\right)}^3={\left(R_1{R}_2\right)}^3=1,\\ {\left(R_2{R}_1{R}_3{R}_1\right)}^4=1,\\ {\left(R_4{R}_5\right)}^3=1\text{.}\\ R_5^2={\left(R_5{R}_2\right)}^2={\left(R_5{R}_1{R}_3{R}_1\right)}^2={\left(R_5{R}_3\right)}^4=R_1\left(R_5{R}_3{R}_2{R}_3\right)R_1{\left(R_5{R}_3{R}_2{R}_3\right)}^{-1}=1\\ R_4^2={\left(R_4{R}_1\right)}^2={\left(R_4{R}_3\right)}^2={\left(R_4{R}_2\right)}^3=1\end{array}$$ It follows from $R_1\left(R_5{R}_3{R}_2{R}_3\right)R_1{\left(R_5{R}_3{R}_2{R}_3\right)}^{-1}=R_5^2=1$ that $$R_5{R}_1{R}_5\in K\prime \text{.}$$

Therefore $$\mathrm{\Gamma }\ni \lambda \left(R_1{R}_5{R}_1{R}_5\right)e_5=\lambda ((1+i)e_1+2e_2+2e_3)$$ and hence $\lambda =\frac{a+bi}{2},$ $a,b\in \mathbb{Z},$ and $a\equiv b$ (mod $2\text{).}$ But $[1,i]e_5\in \mathrm{\Gamma }\text{.}$ Therefore, we may assume that $$\lambda =\frac{1+i}{2}\text{.}$$

It only remains to check whether $c\left(F\right)\in \mathrm{\Gamma }$ or not for this particular $\lambda \text{.}$ This has to be done by straightforward computations: $$\begin{array}{ccc}c\left(r_5^2\right)& =& c\left(r_5\right)+r_{5}c\left(r_5\right)=\frac{1+i}{2}(1+r_5)e=0\in \mathrm{\Gamma }\text{.}\\ c\left({\left(r_4{r}_5\right)}^3\right)& =& (1+r_4{r}_5+{\left(r_4{r}_5\right)}^2)(c\left(r_4\right)+r_{4}c\left(r_5\right))\\ & =& \frac{1+i}{2}(1+r_4{r}_5+{\left(r_4{r}_5\right)}^2)\left(r_4{e}_5\right)=0\in \mathrm{\Gamma },\end{array}$$ and so on (one only needs to consider the relations which involve $R_5\text{).}$ This check shows that $\lambda =\frac{1+i}{2}$ gives a cocycle of $K,$ indeed, and, hence, defines an $r\text{-group}$ $W$ with $\text{Lin}\hspace{0.17em}W=K,$ $\text{Tran}\hspace{0.17em}W=\mathrm{\Gamma }\text{.}$ It can be proved in a straightforward manner that this cocycle $c$ is not a coboundary, i.e. $W$ is not a semidirect product.

The same considerations can be given for each $K$ with $s=n+1,$ and, as a result, one obtains Table 2. It appears a posteriori that in all cases either $\mathrm{\Gamma }$ is a root lattice or $c\left(F\right)\in {\mathrm{\Gamma }}^0$ for every relation $F$ of $K$ (therefore, for an $r\text{-group}$ $W,$ $\text{Tran}\hspace{0.17em}W$ is always a root lattice). By performing similar straightforward computations in the remaining cases, a proof of the theorems in Section 2.8. is obtained.

As for the proof of the theorem from Section 2.9, the first part of it is given in Section 4.6; the second part (about minimality of $\mathbb{Z}\left[\text{Tr}K\right]\text{)}$ follows from [BSh1963]. The third part follows from Table 1.

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